Magnesium powder reacts with steam to form magnesium hydroxide and hydrogen gas.
a) Write a balanced chemical equation for this reaction.
b) What is the percentage yield if 10.1g Mg reacts with an excess of water and 21.0g Mg(OH)2 is recovered?
c) If 24g Mg is used and the percentage yield is 95%, how many grams of magnesium hydroxide should be recovered?
a) Mg + 2H2O ==> Mg(OH)2 + H2
For the reaction shown, calculate how many grams of oxygen form when each quantity of reactant completely reacts.
2HgO(s)→2Hg(l)+O2(g)
a) To write a balanced chemical equation, we need to ensure that the number of atoms of each element is the same on both sides of the equation.
The equation for the reaction between magnesium powder and steam to form magnesium hydroxide and hydrogen gas can be written as follows:
Mg + 2H₂O → Mg(OH)₂ + H₂
b) To find the percentage yield, we need to compare the actual yield (the amount of Mg(OH)₂ recovered) to the theoretical yield (the amount of Mg(OH)₂ that should be obtained based on the stoichiometry of the reaction).
First, we need to calculate the theoretical yield of Mg(OH)₂:
1 mole of Mg reacts with 2 moles of H₂O to produce 1 mole of Mg(OH)₂. The molar mass of Mg(OH)₂ is 58.32 g/mol.
10.1 g of Mg is used, so we can find the moles of Mg:
moles of Mg = mass of Mg / molar mass of Mg
moles of Mg = 10.1 g / 24.31 g/mol = 0.4156 mol
Since the stoichiometric ratio between Mg and Mg(OH)₂ is 1:1, the moles of Mg(OH)₂ produced will be the same as the moles of Mg used.
The molar mass of Mg(OH)₂ is 58.32 g/mol, so the theoretical yield of Mg(OH)₂ is:
theoretical yield = moles of Mg(OH)₂ * molar mass of Mg(OH)₂
theoretical yield = 0.4156 mol * 58.32 g/mol = 24.24 g
Now, we can calculate the percentage yield:
percentage yield = (actual yield / theoretical yield) * 100
percentage yield = (21.0 g / 24.24 g) * 100 ≈ 86.5%
Therefore, the percentage yield is approximately 86.5%.
c) If 24 g of Mg is used and the percentage yield is 95%, we can again calculate the theoretical yield of Mg(OH)₂:
moles of Mg = mass of Mg / molar mass of Mg
moles of Mg = 24 g / 24.31 g/mol = 0.987 mol
theoretical yield = moles of Mg(OH)₂ * molar mass of Mg(OH)₂
theoretical yield = 0.987 mol * 58.32 g/mol = 57.4 g
Now, we can calculate the actual yield based on the given percentage yield:
actual yield = (percentage yield / 100) * theoretical yield
actual yield = (95 / 100) * 57.4 g = 54.53 g
Therefore, if 24 g of Mg is used and the percentage yield is 95%, the expected mass of Mg(OH)₂ to be recovered is approximately 54.53 g.
The equation by Jake is correct. Here is a solved example for stoichiometry problems as well as percent yield. Just follow the steps.
http://www.jiskha.com/science/chemistry/stoichiometry.html